I agree with the series approach but the final result depends on
division which has limited accuracy. I would suggest using the following
which uses 100*L%R as a rational fraction
(10^_2)*(1r2)* 175r637816300
1.37187e_9
However, the solution is wrong in any case. It is the result of using a
circle and a line parallel to the tangent line at any point. In other
words, looking at the earth from a great distance.
The right answer (for a perfect sphere) is 0 as the curvature of the
earth is the same in every direction from the observer. There is no
"arch" It seems that the old Greeks had it right- as ships went off
into the distance, they didn't just shrink and remain whole but would
also disappear gradually from the bottom up. What Patrick Harrington
said is at the core of what is actually observed.
Don Kelly
On 2018-02-26 10:10 PM, J. Patrick Harrington wrote:
The way to avoid loss of accuracy is to expand in a binomial series:
D = R - √(R²-L²) = R[ 1 - (1 - (L/R)^2)^(1/2)]
= R[ 1 - (1 -(1/2)(L/R)^2 + ...)] = (1/2) L*L/R
and thus
]D=. -:L*L%R
1.76384e_9
On Tue, 27 Feb 2018, Ian Clark wrote:
In "The Curious Incident of the Dog in the Night-Time", Mark Haddon
has his
young autistic hero verify that the earth is not flat by holding up a
steel
straight-edge to the horizon.
In the course of private research into public gullibility, I tried to
replicate Haddon's thought-experiment – and failed! Try as I might, I
could
not see the slightest discrepancy between the horizon and my
straight-edge.
I strongly suspect that Haddon hasn't performed the experiment
himself, but
simply written down what he thought one "ought" to see.
So what am I to conclude?
(a) The earth is flat.
(b) The curvature is too small to be seen this way.
(c) Don't believe everything you read in novels.
Let's go with (b), and try to calculate the height D of the "arch" of
the
horizon over the ruler's edge, and show it's too small to observe by the
naked eye.
There's a diagram at:
https://whitbywriters.wordpress.com/2018/02/26/the-curious-incident-of-the-dead-flat-horizon
where I discuss the matter for non-mathematical readers. There I baldly
state that D is given by the formula:
D = R * (1 – cos arcsin L/R)
where L is the half-length of the straight-edge and R the radius of the
earth.
Have I got it right?
I calculated D = 1.4 nm, but omit to say how. (I used J of course).
R=: 6378163 NB. equatorial radius of earth (m)
L=: 0.15 NB. half-length of metal ruler (m)
cos=: 2&o.
arcsin=: _1&o.
smoutput D=: R * (1 - cos arcsin L%R)
1.41624e_9
Now 1.41624e_9 m is roughly 1.4 nm, less than the width of a DNA
molecule.
But I suspect that my estimate of D is unreliable. Moreover it could
be out
by several orders of magnitude. Why? Because I'm calculating the
arcsin of
a very small number, and then calculating the cosine of the resulting
very
small angle, which may be pushing (o.) beyond its limits.
Now I could use Pythagoras instead of trigonometry:
D = R - √(R²-L²)
which avoids using (o.) and may perform the entire calculation in
extended
precision without conversion to (float). I'll also work in nanometres
(nm),
not metres (m):
R=: 6378163000000000x NB. (nm)
L=: 150000000x NB. (nm)
smoutput D=: R - %:(R*R)-(L*L)
2
I thought I was getting 6-figure accuracy, so this result worries me.
2 nm differs significantly from the result of the trig formula: 1.4 nm.
Now (R*R)-(L*L) is extended precision, but (%:) is returning (float) not
(extended). Is it also corrupting the result? Is "2" just an artefact of
its algorithm?
So let's work in picometres, to see if the "2" holds up under greater
precision:
R=: 6378163000000000000x
L=: 150000000000x
smoutput D=: R - %:(R*R)-(L*L)
2048
It does. This is encouraging. But (%:) still returns (float).
Let's try different ways of calculating the square root of an extended
precision result:
sq=: ^&2
sqr=: %:
smoutput D=: R - sqr (sq R)-(sq L)
2048
sqr=: ^&0.5
smoutput D=: R - sqr (sq R)-(sq L)
2048
sqr=: ^&1r2
smoutput D=: R - sqr (sq R)-(sq L)
3072
…which is even more worrying!
So which is the most dependable approximation to D?
3 nm
2 nm
1.4 nm
none of the above?
Is it robust enough to warrant my original deduction?
Have I made a logical error somewhere?
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